Section 12.4: Tangent Planes and Differentials
Let S be the surface defined by z = f (x, y), where f has continuous first partial derivatives
fx and fy . For each point P = (x0 , y0 , z0 ) on the surface, the vertical planes x = x0 and
y = y0 intersect S in curves C1 and C2 . Let T1 and T2 denote the tangent lines to C1 and
C2 at the point P . Then the plane defined by T1 and T2 is called the tangent plane to the
surface S at the point P .
Figure 1: Illustration of the tangent plane to a surface.
Theorem: (Equation of the Tangent Plane)
An equation of the tangent plane to the surface z = f (x, y) at (x0 , y0 , z0 ) is
z − z0 = fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ).
Example: Find an equation of the tangent plane to the surface z = x2 + y 2 + sin(xy) at
(0, 2, 4).
Let f (x, y) = x2 + y 2 + sin(xy). The partial derivatives are
fx (x, y) = 2x + y cos(xy)
fy (x, y) = 2y + x cos(xy).
Then (0, 2, 4), fx (0, 2) = 2 and fy (0, 2) = 4. An equation of the tangent plane is
2(x − 0) + 4(y − 2) = z − 4
2x + 4y − z = 4.
Definition: Consider a function of two variables z = f (x, y). If x and y are given increments
∆x and ∆y, then the corresponding increment of z is
∆z = f (x + ∆x, y + ∆y) − f (x, y).
Thus, the increment ∆z represents the change in the value of z when (x, y) changes to
(x + ∆x, y + ∆y).
Definition: Suppose z = f (x, y). If dx and dy are the differentials of x and y, respectively,
then the differential of z or total differential is given by
dz = fx (x, y)dx + fy (x, y)dy.
Note: If ∆x and ∆y are small, then ∆z ≈ dz. This allows us to estimate the value of
f (x0 + ∆x, y0 + ∆y) when f (x0 , y0 ) is known:
f (x0 + ∆x, y0 + ∆y) ≈ f (x0 , y0 ) + dz.
Example: Let z = 2x2 + 3xy + y 2 . Compare the values of ∆z and dz as (x, y) changes from
(2, 3) to (2.05, 2.96).
The total differential is
dz = (4x + 3y)dx + (3x + 2y)dy.
Setting x = 2, y = 3, dx = ∆x = 0.05, and dy = ∆y = −0.04, we have
dz = (8 + 9)(0.05) − (6 + 6)(0.04) = 0.37,
∆z = f (2.05, 2.96) − f (2, 3) = 0.3706.
Note that ∆z ≈ dz, but dz is easier to compute.
Example: Use differentials to approximate
p
20 − 1.952 − 7(1.08)2 .
Let z = f (x, y) =
p
20 − x2 − 7y 2 . The partial derivatives of f are
x
fx (x, y) = − p
20 − x2 − 7y 2
and
7y
fy (x, y) = − p
.
20 − x2 − 7y 2
Setting x = 2, y = 1, dx = ∆x = −0.05, and dy = ∆y = 0.08, we have
p
20 − 1.952 − 7(1.08)2 = f (1.95, 1.08)
≈ f (2, 1) + dz
= f (2, 1) + fx (2, 1)dx + fy (2, 1)dy
2
7
= 3 − (−0.05) − (0.08)
3
3
= 2.8467.
Example: The base radius and height of a right circular cone are measured as 10 cm and
25 cm, respectively, with a possible error in measurement of at most 0.1 cm in each. Use
differentials to estimate the maximum error in the calculated volume of the cone.
The volume of a cone with base radius r and height h is
V =
π 2
r h.
3
So the differential of V is
dV =
∂V
∂V
2
π
dr +
dh = rhdr + r2 dh.
∂r
∂h
3
3
Since the errors are at most 0.1 cm, |∆r| ≤ 0.1 and |∆h| ≤ 0.1. Setting r = 10, h = 25, and
dr = dh = 0.1, we have
dV =
100π
500π
(0.1) +
(0.1) = 20π.
3
3
Thus, the maximum error in the calculated volume is approximately 20π ≈ 63 cm3 .
Note: The definitions of an increment or total differential can be extended to functions of
three or more variables. If w = f (x, y, z), then the increment of w is
∆w = f (x + ∆x, y + ∆y, z + ∆z) − f (x, y, z),
and the differential of w is
dw = fx (x, y, z)dx + fy (x, y, z)dy + fz (x, y, z)dz.